On lattices, modules and groups with many uniform elements
The uniform dimension, also known as Goldie dimension, can be defined and used not only in the class of modules, but also in large classes of lattices and groups. For considering this dimension it is necessary to involve uniform elements. In this paper we are going to discuss properties of latti...
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irk-123456789-1559512019-06-18T01:30:59Z On lattices, modules and groups with many uniform elements Krempa, J. The uniform dimension, also known as Goldie dimension, can be defined and used not only in the class of modules, but also in large classes of lattices and groups. For considering this dimension it is necessary to involve uniform elements. In this paper we are going to discuss properties of lattices with many uniform elements. Further, we examine these properties in the case of lattices of submodules and of subgroups. We also formulate some questions related to the subject of this note. 2004 Article On lattices, modules and groups with many uniform elements / J. Krempa // Algebra and Discrete Mathematics. — 2004. — Vol. 3, № 1. — С. 75–86. — Бібліогр.: 18 назв. — англ. 1726-3255 2000 Mathematics Subject Classification: 06C99, 16S90, 20E15. http://dspace.nbuv.gov.ua/handle/123456789/155951 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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The uniform dimension, also known as Goldie
dimension, can be defined and used not only in the class of modules,
but also in large classes of lattices and groups. For considering this
dimension it is necessary to involve uniform elements.
In this paper we are going to discuss properties of lattices with
many uniform elements. Further, we examine these properties in
the case of lattices of submodules and of subgroups. We also formulate some questions related to the subject of this note. |
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Krempa, J. |
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Krempa, J. On lattices, modules and groups with many uniform elements Algebra and Discrete Mathematics |
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Krempa, J. |
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Krempa, J. |
title |
On lattices, modules and groups with many uniform elements |
title_short |
On lattices, modules and groups with many uniform elements |
title_full |
On lattices, modules and groups with many uniform elements |
title_fullStr |
On lattices, modules and groups with many uniform elements |
title_full_unstemmed |
On lattices, modules and groups with many uniform elements |
title_sort |
on lattices, modules and groups with many uniform elements |
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Інститут прикладної математики і механіки НАН України |
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2004 |
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http://dspace.nbuv.gov.ua/handle/123456789/155951 |
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On lattices, modules and groups with many uniform elements / J. Krempa // Algebra and Discrete Mathematics. — 2004. — Vol. 3, № 1. — С. 75–86. — Бібліогр.: 18 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT krempaj onlatticesmodulesandgroupswithmanyuniformelements |
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2025-07-14T08:09:16Z |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 1. (2004). pp. 75 – 86
c© Journal “Algebra and Discrete Mathematics”
On lattices, modules and groups with many
uniform elements
Jan Krempa
Communicated by Z. S. Marciniak
Abstract. The uniform dimension, also known as Goldie
dimension, can be defined and used not only in the class of modules,
but also in large classes of lattices and groups. For considering this
dimension it is necessary to involve uniform elements.
In this paper we are going to discuss properties of lattices with
many uniform elements. Further, we examine these properties in
the case of lattices of submodules and of subgroups. We also for-
mulate some questions related to the subject of this note.
1. Preliminaries
In this section we present basic notions and results on lattices. For con-
venience we assume that all lattices have 0 and 1. Sublattices with the
same 0 will often be called 0-sublattices. Further notation and termino-
logy on lattices is similar to that from [6]. However, as in the case of
submodules and subgroups, in any lattice L, if a ≤ b then the interval
[a, b] will be denoted by [b/a]. Intervals of the form [a/0] will often be
called 0-intervals. Our crucial examples are:
Example 1.1. Let R be an associative ring with unity, M a (right) R-
module and L(M) the set of all submodules of M ordered by inclusion.
It is known that L(M) is an algebraic and modular lattice with 0 = 0R
and 1 = MR.
Many thanks to Piotr Grzeszczuk for his helpful remarks on preliminary version
of Section 3 of this paper.
2000 Mathematics Subject Classification: 06C99, 16S90, 20E15.
Key words and phrases: uniform element, locally uniform lattice, lattice of
subgroups, lattice of submodules.
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.76 On uniform elements
Example 1.2. Let G be a group and L(G) be the set of all subgroups of
G ordered by inclusion. Then it is well known that L(G) is an algebraic
lattice with 0 = 〈e〉 and 1 = G. This lattice need not be modular.
The above examples and study of uniform dimension suggest to con-
sider not only modular lattices. Thus some notions, defined in fact in
[12, 18, 11] are of interest here. A lattice L is balanced if L satisfies the
following quasi-identity:
(x ∧ y) ∨ ((x ∨ y) ∧ z) = 0 =⇒ (y ∨ z) ∧ x = 0
and L is strongly balanced if all nonempty intervals of L are balanced lat-
tices. The last property can be expressed by the following quasi-identity:
(x ∧ y) ∨ ((x ∨ y) ∧ z) = x ∧ y ∧ z =⇒ (y ∨ z) ∧ x = x ∧ y ∧ z.
By definition we have that strongly balanced lattices form a quasi-variety
of all lattices and balanced lattices form a quasi-variety of lattices with 0.
Some properties of these quasi-varieties can be found for example in [12,
11, 18]. For us it is important to remember that any modular lattice
is strongly balanced and any strongly balanced lattice is balanced, but
need not be modular: the pentagon is the smallest nonmodular, strongly
balanced lattice. The smallest example of a nonbalanced lattice is the left
picture in Figure 1, while the smallest balanced but not strongly balanced
lattice is the right picture in Figure 1.
Let L be a lattice and a ∈ L. Then, as in [8, 7, 12], a is called essential
(in L) if for any b ∈ L from a ∧ b = 0 it follows b = 0. Also, b ∈ L is
an E-complement of a (in L) if a ∧ b = 0 and a ∨ b is essential in L.
Further, L will be called E-complemented if for any a ∈ L there exists an
E-complement of a in L. These notions are of special interest in balanced
lattices. For example, from [7, 12] we have:
Proposition 1.3. Let L be a balanced lattice.
• Any pseudocomplement in L is an E-complement.
• If L is algebraic then L is E-complemented.
• Let L be E-complemented and a, b, c ∈ L are such that a∨b ≤ c and
a ∧ b = 0. Then there exists an E-complement b′ of a in [c/0] such
that b ≤ b′. In particular, the interval [c/0] is E-complemented.
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.J. Krempa 77
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Figure 1: Some small lattices
Note here that in the left lattice from Figure 1 the element b is a
pseudo-complement of a, but it is not an E-complement. The following
question seems to be not very difficult.
Question 1. Let L be a lattice satisfying the following condition: For any
c ∈ L and a, b ∈ [c/0] such that a ∧ b = 0 there exists an E-complement
b′ of a in [c/0] with b ≤ b′. Is L a balanced lattice?
The Ascending (Descending) Chain Condition for any partially or-
dered set will be abbreviated by ACC (DCC respectively). If L and
M are lattices, then trivial extension of L by M will mean here the set
L∪ (M \{0}) with the relations ≤ from L and M and with l < m for any
l ∈ L, m ∈ M \ {0} (see pictures on Figure 1 as a simple illustration).
2. Uniform elements
In this section lattices having many uniform elements will be considered.
Recall, following [13, 8, 12], that if L is a lattice and u ∈ L then u
is uniform (in L) if u 6= 0 and any 0 6= x ∈ [u/0] is essential in this
interval. Moreover, L is uniform if 0 6= 1 and 1 is a uniform element
in L. Obviously an element u ∈ L is uniform if and only if the interval
[u/0] is a uniform lattice and any nonzero element of a uniform lattice is
uniform. In particular atoms are uniform elements.
It is evident that any chain is a uniform lattice and any uniform lattice
is balanced, but it need not be strongly balanced.
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.78 On uniform elements
Example 2.1. Let L be a uniform lattice and let M be a nonbalanced
lattice. Then the trivial extension of L by M is a uniform, but not
a strongly balanced lattice.
As in [14, 12], a lattice L will be named locally uniform if any 0-
interval in L contains a uniform element. Thus every atomic lattice, in
particular any finite lattice, is locally uniform.
Clearly any 0-interval of locally uniform lattice is locally uniform and
any lattice with DCC for elements is locally uniform too. On the other
hand, in [10, 9, 16] there are examples of 0-sublattices of distributive
locally uniform lattices, being not locally uniform. We also have
Proposition 2.2. Let L be an E-complemented balanced lattice with ACC
for E-complements. Then L is locally uniform.
Proof. Let 0 6= b ∈ L. We have to find a uniform element in [b/0].
From Proposition 1.3 and the assumption we know that this interval
is E-complemented. Hence, we can assume that b = 1 and [b/0] = L.
If L is uniform then there is nothing to prove. Let L be not uniform.
Denote by X the set of all E-complements of nonzero elements. By as-
sumption this set is not empty and contains a maximal element, say c.
Hence there exists 0 6= a ∈ L such that a ∧ c = 0 and a ∨ c is essen-
tial in L. Let d, e ∈ [a/0] be such that d 6= 0 and d ∧ e = 0. Evidently
(d∨e)∧c = 0. Thus, by balancedness of L, we have d∧(e∨c) = 0. Hence,
by Proposition 1.3, there exists an E-complement f of d in L such that
e ∨ c ≤ f. Hence f ∈ X. Now maximality of c in X implies that f = c.
Thus e = 0, because c ∧ e = 0 and e ≤ c. Hence a has to be uniform, as
required.
The above result can, by Proposition 1.3, be applied to balanced lat-
tices with ACC for elements. However the assumption on balancedness is
indispensable. Indeed, in [10] a lattice with ACC for elements, but having
no uniform elements is constructed. It is also known (see Proposition 3.5
below) that without the ACC condition the above proposition need not
be true for modular algebraic lattices.
One might expect that in any algebraic lattice the interval generated
by all uniform elements is locally uniform, but it is not true, even for
lattices of submodules (see Example 3.9 below). Only under very strong
conditions this is true.
Proposition 2.3. Let L be an algebraic and distributive lattice and let
a ∈ L be the join of all uniform elements. Then the interval [a/0] is a
locally uniform lattice.
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Proof. Let 0 6= b ∈ [a/0]. We can assume that b is compact. Thus, by
definition of a, we can assume that there are uniform elements u1, . . . , un
such that b ≤
∨n
i=1
ui. Hence, by distributivity, b =
∨n
i=1
(b ∧ ui). This
means that for some i the meet b∧ ui is nonzero and is uniform, because
it is smaller than ui.
One can observe the following fact about extensions, covering the case
of trivial extensions:
Proposition 2.4. Let L be a lattice and a ∈ L.
• If a is essential in L and the interval [a/0] is locally uniform then
L is locally uniform;
• If a is a modular element and the intervals [a/0] and [1/a] are locally
uniform then L is locally uniform.
As in [16], a lattice L will be called strongly locally uniform if any its
nontrivial interval is locally uniform. Thus every strongly atomic lattice
is strongly locally uniform. Hence, lattices satisfying DCC for elements
are strongly locally uniform. Chains form another class of strongly locally
uniform lattices.
Immediately from the definition any interval of a strongly locally uni-
form lattice is strongly locally uniform. On the other hand, in [10, 9, 16]
there are examples of distributive, strongly locally uniform lattices con-
taining 0-sublattices being even not uniform. From [10] we have:
Proposition 2.5. Let L be a lattice with ACC for elements. If L is
strongly balanced then L is strongly locally uniform.
Lemma 2.6. Let L be a lattice and a ∈ L be an element such that any
b ∈ [a/0] is modular in L. If the intervals [a/0] and [1/a] are strongly
locally uniform then L is strongly locally uniform.
As a consequence we obtain
Theorem 2.7. Let L be a complete and modular lattice. Then there exists
an element a ∈ L such that the interval [a/0] is strongly locally uniform
and contains all strongly locally uniform 0-intervals of L. Moreover, the
lattice [1/a] has no strongly locally uniform interval of the form [b/a].
Proof. Clearly the union of any increasing chain of strongly locally uni-
form 0-intervals in L is strongly locally uniform. Hence, by Zorn’s Lemma,
we can take a maximal interval, say [a/0], of this type. Using Lemma 2.6
one can check that the interval [1/a] has no nontrivial strongly locally
uniform intervals of the form [b/a].
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.80 On uniform elements
Let c ∈ L be such that the interval [c/0] is strongly locally uniform.
Then, by definition, the interval [c/(a∧c)] is also strongly locally uniform.
Hence, by modularity of L, the interval [(a ∨ c)/a] is strongly locally
uniform too. By previous part of the proof we then have c ≤ a, as
required.
Question 2. Can the above result be proved for strongly balanced alge-
braic lattices?
Our interest to study lattices, modules and groups with many uniform
elements was motivated by the following result from [9, 12, 11]:
Theorem 2.8. Let L be a lattice. Then the (strong) uniform dimension
is defined for L if and only if L is (strongly) balanced and (strongly) locally
uniform.
3. Modules
In this section we discuss modules with lattices of submodules having
many uniform elements. For our convenience modules will often be named
according to properties of lattices of their submodules.
There exists a rich theory of chain modules and of uniform modules,
often depending on particular properties of the basic ring (see [13]). In-
jective hulls are very important in this theory. In particular we have
Proposition 3.1. Let M be a module with the injective hull M. Then M
is uniform or locally uniform if and only if M shares the same property.
From results of previous sections and modularity of lattices of the form
L(M) we have that a submodule of a (strongly) locally uniform module is
(strongly) locally uniform and homomorphic image of a strongly locally
uniform module is strongly locally uniform. However, (see Example 3.9
below), a homomorphic image of a locally uniform module need not be
locally uniform. We only have
Proposition 3.2. For any ring R the class of all locally uniform R-
modules is closed under direct sums. Thus, there exists a locally uniform
and faithful R-module.
Proof. Let M1, M2 be locally uniform R-modules and 0 6= X ⊆ M1 ⊕M2
be a submodule. If X ∩ M1 = 0 then X contains a uniform submodule
isomorphic to a submodule of M2. If X ∩ M1 6= 0 then this intersection,
and X too, contains a uniform submodule. Hence M1 ⊕ M2 is locally
uniform. The same conclusion follows easily for direct sums of any family
of locally uniform modules.
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Now, for any 0 6= r ∈ R let Ir be a maximal right ideal of R not
containing r, and Mr = R/Ir. Then it is easy to check that all modules
Mr are uniform and the module M =
⊕
r∈R Mr is a faithful R-module.
By the first part of the proof M is locally uniform.
We also have
Proposition 3.3. Let R be a ring. Then the following conditions are
equivalent:
1. Any R-module is locally uniform;
2. Any R-module is strongly locally uniform;
3. Any cyclic R-module is locally uniform;
4. Any nonzero injective R-module contains a nonzero injective inde-
composable submodule.
Proof. Let M be an R-module. It is evident that any interval in the
lattice L(M) is isomorphic to L(N) for a homomorphic image N of a
submodule of M. Hence, if any R-module is locally uniform then any
R-module is strongly locally uniform. Now the result follows by standard
arguments.
Let us call a ring R an LU-ring if any R-module is locally uniform.
Certainly right noetherian rings are LU-rings. More generally, combining
Corollary 2.10 from [4] with Proposition 2.6(i) in the same paper we
obtain the following result, noticed in [9]
Theorem 3.4. Any ring with right Gabriel dimension is an LU-ring.
On the other hand we have the following result, in fact due to Goldie
Theorem 3.5. Let R be a domain. Then the following conditions are
equivalent:
1. The module RR is uniform;
2. The module RR is locally uniform;
3. The module RR contains a uniform submodule;
4. R is a right Ore domain.
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.82 On uniform elements
Proof. Let the module RR contain a uniform submodule X and let 0 6=
x ∈ X. Then the module xR ⊆ X is uniform and isomorphic to RR,
because R is a domain. In this way 3 ⇒ 1. Now it is easy to complete
the proof.
From the above result we know that any commutative domain is an
LU-ring, but there are examples of commutative semiprime rings without
this property. It is enough to take a commutative regular (in the sense
of von Neumann) ring without minimal idempotents, as in Example 2 in
[9].
As a consequence of results from [5] we know that any right chain ring
has right Krull dimension if and only if it has right Gabriel dimension.
Hence in [4, Example 10,10] it is shown in fact an example of a com-
mutative local valuation domain R without Gabriel dimension. However
this domain is an LU-ring by Proposition 3.3. Using the same arguments
one can see that any homomorphic image of R, different from the residue
field, is also an LU-ring without Gabriel dimension, but has only one
prime ideal. Thus it would be interesting to find an intrinsic characteri-
zation of the class of all LU-rings, not covered by Proposition 3.3. Results
saying that some well known classes of rings contain only LU-rings are
also of interest.
For arbitrary ring, using Theorem 2.7, we obtain:
Theorem 3.6. Let M be a module. Then there exists the largest submod-
ule S(M) which is strongly locally uniform. Moreover, S(M/S(M)) = 0.
As a consequence of the above result and earlier observations we have
Corollary 3.7. Let R be a ring. Then the class of all strongly locally
uniform modules is closed under unions of chains, taking submodules,
homomorphic images and extensions.
The above corollary means that for any ring R to be strongly locally
uniform is a hereditary torsion in the category of all R-modules. This
torsion is certainly nonzero, because there are simple R-modules, but it
can be proper.
Example 3.8. Let F be a field and let A be an F -algebra which is
a domain but not a right Ore domain, for example the free associative
algebra with more than one generator. Let V be the algebra A considered
as a vector space over F and let E = EndF (V ). Then A ⊂ E by right
multiplications. Let B ⊂ E be the set of all endomorphisms of finite rank
and R = A + B. Then it is known, and easy, that B is an ideal essential
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in L(RR). Moreover, B is the right socle of R and R/B ≃ A. Hence, by
Theorem 3.5, S(RR) = B.
Now let W ⊂ V be a subspace of codimension 1 and let M = {r ∈
R | Wr = 0}. Certainly M ⊂ B and M is a maximal submodule of B.
Let X = R/M. Then one can check that X is uniform, because B/M is
an essential submodule in X, but is not strongly locally uniform, because
X/(B/M) ≃ R/B. In this way we obtained that S(X) = B/M 6= X.
Now we show that a module generated by two uniform submodules
need not be locally uniform.
Example 3.9. Let X be a uniform module, and let P ⊂ X be a submod-
ule such that X/P is not locally uniform. Let N = {(p, p) | p ∈ P} and
Y = (X × X)/N. Then Y is generated by uniform submodules, natural
images of X × 0 and 0 × X. On the other hand, Y contains a natural
image of X embedded diagonally into X × X. This image is isomorphic
to X/P and is not locally uniform. Hence Y is not locally uniform.
In connection with just indicated examples the following question
seems to be interesting
Question 3. Let R be a ring, but not an LU-ring. Is there a uniform
R-module M with S(M) = 0?
4. Groups
In this section groups with lattices of subgroups having a lot of uniform
elements are discussed. Results exhibiting difference from lattices of the
form L(M) will be of special interest. Groups will often be named ac-
cording to properties of lattices of their subgroups.
It is well known ([17]) that a group G is a chain if and only if there
exists a prime p such that G is a subgroup of the Prüfer group Cp∞ .
It is evident that any uniform group is either a p-group or is torsion
free. However the structure of all uniform groups is complicated and is
not known yet. To discuss details let us agree that the following groups
are standard uniform groups:
• subgroups of Prüfer groups Cp∞ for all primes p,
• subgroups of the additive group Q of the rational numbers,
• subgroups of the infinite quaternion group Q2∞ .
It is visible that any standard uniform group has an abelian subgroup
of index at most 2. Hence, it would be interesting to know if uniform
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groups, maybe with some finiteness conditions, have to be standard. In
this direction, among others, the following results are proved in [1, 10]:
Theorem 4.1. Let G be a locally virtually solvable group. If G is uniform
then G is a standard uniform group. In particular, any locally linear
uniform group is standard uniform.
Proposition 4.2. Let G be a periodic locally residually finite group. If
G is uniform then G is a standard uniform group.
Question 4. Let G be a torsion free and locally residually finite uniform
group. Is G a standard uniform group?
Examples of nonstandard uniform groups were constructed by Adian,
Ol’shanskii and others. Some of these examples are presented in [15, §31].
Let us recall the notion invented (under another name) over 30 years
ago by S.N. Černikov in [2]: A group G is locally graded if any finitely
generated subgroup 〈e〉 < H ≤ G contains a subgroup of finite index. In
connection with the above results the following question is of interest:
Question 5. Let G be a locally graded uniform group. Is G a standard
uniform group?
Any cyclic group is certainly locally uniform. Hence we have the
following immediate observation
Proposition 4.3. Let G be a group. Then G is locally uniform.
Situation with strongly locally uniform groups is complicated. Cer-
tainly any abelian group is strongly locally uniform, but in [10] there are
examples of metabelian, but not strongly locally uniform groups. Among
them are locally finite groups. Thus we know that an extension of a
strongly locally uniform group by another strongly locally uniform group
need not be strongly locally uniform.
In [10, 16] some strongly locally uniform groups are described:
Theorem 4.4. Let G be a group. Then G is strongly locally uniform in
any of the following cases:
1. G satisfies DCC for subgroups,
2. G is strongly balanced and satisfies ACC for subgroups,
3. G is strongly balanced and locally finite,
4. G satisfies the normalizer condition,
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5. G is generalized nilpotent.
In [16] it was shown that some cyclic extensions of periodic abelian
groups are strongly locally uniform, and even strongly atomic. As an
immediate consequence of Lemma 2.6 we have a related result.
Theorem 4.5. Let G be a group and N ⊆ G a normal subgroup such
that every subgroup of N is modular in G. If N and G/N are strongly
locally uniform groups, then G is also strongly locally uniform.
Corollary 4.6. Let N ⊆ G be a subgroup such that any subgroup of N is
normal in G. If G/N is strongly locally uniform then G is strongly locally
uniform too.
Question 6. Let G be a strongly balanced group. Is G strongly locally
uniform?
For locally finite groups the affirmative answer follows from [16]. On
the other hand, by examples from [10] we know that for arbitrary lattices
the answer is in the negative.
References
[1] C. Bagiński and J. Krempa, On a characterization of infinite cyclic groups, Publ.
Math. Debrecen 63(2003), 249-254.
[2] S.N. Černikov, Infinite nonabelian groups with an ivariance condition for infinite
nonabelian subgroups, Dokl. Akad. Nauk SSSR 194(1970), 1280-1283.
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Contact information
J. Krempa Institute of Mathematics
Warsaw University
ul. Banacha 2, 02-097 Warszawa
POLAND
E-Mail: jkrempa@mimuw.edu.pl
Received by the editors: 25.11.2003
and final form in 29.01.2004.
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